Diameter of globe = 30 m
Radius of globe, r = 15 m
Let 'h' be the minimum height from which the cyclist after rolling down an incline will acquire velocity =\(\sqrt{2gh}\)
For looping the loop, the minimum velocity at the lowest point should be\(\sqrt{5gr}\) .
\(\therefore \sqrt{5gh} =\sqrt{2gh}\)
or h = \({5r\over 2}={5\times 15\over 2}=37.5m\)

In solving inclined plane problems, the x and y directions along which the forces are to be considered, may be taken as shown. The components of weight of the box are
(i) mg sin \(\alpha\)acting down the plane and
(ii) mg cos \(\alpha\) acting perpendicular to the plane.

N = mg cos \(\alpha\)
mg sin \(\alpha\) -\(\mu\) N =ma
mg sin \(\alpha\) -\(\mu\) mg cos \(\alpha\) =ma
a = g sin \(\alpha\) -\(\mu\)g cos \(\alpha\)
= g (sin \(\alpha\) - \(\mu\)cos \(\alpha\))
= 9.8 ( \({1\over 2}\) - 0.4 x\({\sqrt{3}\over 2}\))
= 4.9 x 0.3072 = 1.505 m/S²
The box accelerates down the plane at 1.505 m/S²

Consider two bodies A and B, with inital momenta \(\overrightarrow{p}\)_A and \(\overrightarrow{p}\) _B respectively. Let the two bodies collide, get apart and have final momenta\(\overrightarrow{p}\) _A and \(\overrightarrow{p}\) _B respectively. By the second law of motion:
 Change in momentum of body A, \(\overrightarrow{p}\) _A - \(\overrightarrow{p}\)_A =\(\overrightarrow{F}\)_AB \(\triangle t\).......(i)
where \(\overrightarrow{F}\)_AB is the force acting on A due to action of B for a time\(\triangle t\).
Similarly change in momentum of body B, \(\overrightarrow{p}\)B - \(\overrightarrow{p}\)B = \(\overrightarrow{F}\)BA \(\triangle t\) ........(ii)
Here time \(\triangle t\), the time for which two bodies A and B are in contact and interact, is same for both the forces.
Moreover, from third law of motion \(\overrightarrow{F}\)_AB - \(\overrightarrow{F}\)_BA
Hence, adding (i) and (ii), we obtain
(\(\overrightarrow{p}\)_A -\(\overrightarrow{p}\) _A ) + (\(\overrightarrow{p}\)_B - \(\overrightarrow{p}\)_B ) = \(\overrightarrow{F}\)_AB\(\triangle t\)+ \(\overrightarrow{F}\)_BA\(\triangle t\)=- \(\overrightarrow{F}\)_BA\(\triangle t\)+ \(\overrightarrow{F}\)_BA\(\triangle t\)=0
\(\Rightarrow\)\(\overrightarrow{p'}\) _A + \(\overrightarrow{p'}\)_B = \(\overrightarrow{p}\) A + \(\overrightarrow{p}\) B
which shows that the total final momentum of the isolated system is exactly same as its initial momentum. Thus, it is proved that total momentum of an isolated system remains conserved.

\(\because \)  Mass of each coin = m
Number of total coins = 10
Force acting on 7th coin by the 8th coin = weight of the 8 coins + weight of two coins supported by  8 coins.
= mg + 2 mg
= 3 mg N ( downward)

\(\because \)  Mass of the helicopter, m₁ = 1000 kg
Mass of the crew and the passengers, m₂ = 300 kg
Acceleration of the helicopter, a = 15 m/ s²
Acceleration  due to gravity, g = 10 m/s²
(i) Let R, be the reaction applied by the floor on the crew and the passengers.

R₁- m₂g = m₂a
or
R₁= m₂g + m₂a = m₂(g+a)
= 300(10 + 15) = 7500N(upward direction)
(ii) Action of the rotor of the helicopter on the surrounding air
= ( m₁ + m₂ ) g + ( m₁ + m₂ ) a
= ( m₁ + m₂ ) ( g + a )
= ( 1000 + 300 ) x (10 + 15 )
= 1300 x 25 = 32500 N
Force (action ) of the rotor of the helicopter on the surrounding air = 32500 N (downward)

The maximum speed which the car can have without skidding is given by
\(\mu =\frac { { v }^{ 2 } }{ rg } \Rightarrow v=\sqrt { \mu rg }\)
\( \\ Here,\ r=100m,\mu =0.8,g=9.8m{ s }^{ -2 }\)
\(\\ v =\sqrt { 0.8\times 100\times 9.8 } =\sqrt { 4\times 2\times 2\times 49 } \)
\(\\ v\ =\ 2\times 2\times 7\ = 28{ m }/{ s }\)

Given, r =10⁴ km = 10⁴ \(\times \) 1000 m = 10⁷ m,
m = 2500 kg
v = 4 kms^-1 = 4\(\times \)10³ ms^-1
Now,centripeta force is F = \(\frac { m{ v }^{ 2 } }{ r } \) 
F = \(\frac { 2500\times (4\times { 10 }^{ 3 }{ ) }^{ 2 } }{ { 10 }^{ 7 } } =\frac { 2500\times 16\times { 10 }^{ 6 } }{ { 10 }^{ 7 } }\)
\( \\ F=\quad \frac { 250\times 16\times { 10 }^{ 7 } }{ { 10 }^{ 7 } }\)
\(= 4000N\)

Rubber tyres are preferred because coefficient of friction between rubber and road is less than that between steel and road.

Newton's third law of motion

(i) Apparent weight = m (g + a)
(ii) Apparent weight = m (g - a).

The part correctly describes the trajectory of the stone after the spring breaks because when a stone tied to one end of a string is whirled round in a circle then velocity of the stone at any point is along the tangent at that point. If the string breaks suddenly, then stone flies off tangentially, along the direction of its velocity.